Cube Root of x
Show that for any non-zero starting point x0 , Newton’s method will never find
the exact value x for which x3 = 0.
Solution
Given a value x0 close to some x where f (x) = 0, Newton’s method usually
produces a series of values x0 , x1 , x2 , ... whose values approach x. However,
there is no guarantee that any of those values xi actually equal x.
Newton’s method generates a sequence according to the formula:
xk+1 = xk −
f (xk )
.
f � (xk )
In our example, f (x) = x3 and f � (x) = 3x2 , so:
xk+1
xk+1
x3k
3x2k
xk
= xk −
3
2
=
xk
3
xk −
=
Starting with x0 , Newton’s method generates the sequence:
x1
=
x2
=
x3
=
x4
=
2
x0
3
�
�
2
2
22
2
x1 = ·
x 0 = 2 x0
3
3
3
3
2
23
x 2 = 3 x0
3
3
2
24
x 3 = 4 x0
3
3
..
.
xn
=
2n
x0
3n
If x0 =
� 0, the values of the xi get closer and closer to the desired value 0 but
never exactly equal zero.
1
MIT OpenCourseWare
http://ocw.mit.edu
18.01SC Single Variable Calculus��
Fall 2010 ��
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.